Throughout, let $f$ be a Lebesgue measurable function (or continuous if you wish, but this is probably no easier). (Questions with distributions etc. are possible also but I want to keep things simple here).

FINAL CLARIFICATION/REWRITE!!

Thanks to all who have commented so far. I will need some more time to digest it properly. The original forms of the questions are at the end; here I have rewritten the questions, hopefully more clearly; sorry for my poor explanation before!

Thus, trivially $0 \in \mathcal{M}_a \subseteq \mathcal{M}$, but $\mathcal{M}_a$ contains many other non-trivial functions. It seems certain that $\mathcal{M}_a \ne \mathcal{M}$ (I would be amazed if the spaces were equal), although constructing an explicit example seems tricky.

Definition Given a function $\psi \geq 0$, let $G(\psi)$ be the set of all $f$ such that $|f| \leq \psi$.

(Of course we're identifying functions equal a.e., so really we should consider equivalence classes etc. just as for $L^p$ spaces).

Rephrased Question 2 The same as Question 1, but with $G(\psi) \cap \mathcal{M}$ instead.

Or, if $G(\psi)$ is not the appropriate space for these problems, consider $\int_T^{2T} |f| dt \leq g(T) $ or $\int_n^{n+1} |f| dt \leq A_n$ or something similar instead. Finding the correct kind of restrictions on $f$ is part of the problem.

Original QUESTION 1

EDIT: CLARIFICATION: this is really about classes of functions, expressed in terms of a growth/decay rate function $\phi$, which giveuniquesolutions to the moment problem. I am NOT asking how to solve the moment problem itself!

But $\phi = \chi_{[0,b]}$ is enough (by Example 1 below); moreover $\phi(R) = \exp(-\delta R)$ would be enough for any fixed $\delta > 0$, by my discussion of Example 1 below, because the relevant Laplace transform $F = \mathcal{L}f$ is analytic on the half-plane $\{ \mathrm{Re}(z) > -\delta \}$. So we want to know about the gap between $\exp(-\delta R)$ and functions like that given by "coudy" below.

FURTHER CLARIFICATION: by analogy, maybe an example from PDEs will explain better (I'm not saying this is related to my problem; I'm saying that this is the same kind of result as what I want):

Theorem There exists a null temperature function satisfying $|u(x,t)| \leq \exp(A/t)$ with $A>0$, such that $u(x,t) \not\equiv 0$ for all $t>0$.

Theorem Let $u$ be a null temperature function satisfying $|u(x,t)| \leq A \exp(B t^{-\delta})$, for some $A,B>0$ and $\delta<1$. Then $u \equiv 0$.

So here the "critical" growth rate for null temperature functions is roughly $\exp(A/t)$. I am looking for a similar thing with "null moment functions".

Note that this is a totally different problem to: given $v$, find some $u$ satisfying the heat equation such that $u(x,0)=v(x)$.

Original QUESTION 2

If instead we have only $\int_0^R |f(t)| dt < \infty$ for each $R>0$, and
$$
\lim_{R \to \infty} \int_0^R t^n f(t) dt = 0, \qquad n=0,1,2,\ldots
$$
when must $f \equiv 0$ almost everywhere? (I have very little idea about this).

I think these questions are clearly very natural, interesting, and important, but Googling etc. didn't work well (I tried "vanishing moments" and other phrases, but there's just too much stuff out there). Standard known examples/methods follow.

Example 1: if $f$ is compactly supported on $[a,b]$, say, then $f \equiv 0$ a.e. because polynomials are dense in $C[a,b]$.

So, any condition on $f$ forcing $F$ to be analytic on some disc with centre $0$ is enough; but can we do better?

In Example 2, $f \in L^1(0,\infty)$ and so $F$ is bounded and analytic on $\{ \mathrm{Re}(z) > 0 \}$, and continuous on the boundary, with $\lim_{z \to 0} F^{(n)}(z) = 0$ for all $n$. But this is still not enough to force $F \equiv 0$.

Thanks very much! But although this is interesting, I don't see how it works for my problem. I am not asking "given a sequence of numbers, find a function f in a specific class with those moments" - all my moments are zero, so I do know one solution, trivially! I want to know how nasty the others must be; unfortunately I don't know which class of functions is relevant (that's the whole point of my question).
–
Zen HarperJul 10 '10 at 14:46

1

if you suppose a strong rate of decay (sub-gaussian for example), then your statement becomes true. See Terry Tao's blog for a discussion (section on the moment method) on related ideas: terrytao.wordpress.com/2010/01/05/…
–
AlekkJul 10 '10 at 14:49

@Zen. To see how your problem connects to the problem of moments in probability theory, just write f as $f_+-f_-$. Now saying that f has zero moments amounts to saying that the two measures $f_+dx$ and $f_ dx$ have the same moments. Does that imply that the two measures are equal ? The link provided by Mariano may lead you to the [[Carleman's condition][en.wikipedia.org/wiki/Carleman%27s_condition]], which traduces back into a condition on the decreasing rate of the moments of $|f|$.
–
coudyJul 10 '10 at 15:14

Thanks very much! This is interesting, but only a very partial answer to my first question: it shows that a decay rate $\int_R^\infty |f(x)| dx \leq \phi(R)$ with some particular $\phi$ is not enough to force $f=0$. If possible, I would really like a general necessary and sufficient condition on $\phi$.
–
Zen HarperJul 10 '10 at 14:54

Sufficient I can understand, and see Alekk's comment about sub Gaussianity, but what do you mean by necessary? If you mean something like "if $F$ is a class of functions such that only the trivial element $f = 0$ in $F$ has all moment zero, then $F$ must be dominated by some rate $\phi$", I think you won't get anything meaningful. Take for example the class of non-negative, continuous, Lebesgue integrable functions. There are definitely elements in there who decay slower than the Gaussian.
–
Willie WongJul 10 '10 at 15:22

1

I'll accept this since it tells me something interesting and nontrivial I didn't know before, and I feel guilty since my original question is maybe a bit stupid.
–
Zen HarperSep 19 '10 at 10:10